1. 6.4 The Method of Sections
Used to determine the loadings
within a body
If a body is in equilibrium, any part
of the body is in equilibrium
To determine the forces within the
members, an imaginary section
indicated by the blue line, can be
used to cut each member into two
and expose each internal force as
external

2. 6.4 The Method of Sections
It can be seen that equilibrium
requires the member in tension (T) be
subjected to a pull and the member in
compression (C) be subjected to a
push
Method of section can be used to cut
or section members of an entire truss
Apply equations of equilibrium on that
part to determine the members

3. 6.4 The Method of Sections
Consider the truss shown
To determine the force in the member
GC, section aa would be considered

4. 6.4 The Method of Sections
Consider the FBD
Note the line of action of each member force is
specified from the geometry of the truss
Member forces acting on one part of the truss
are equal and opposite to those acting on the
other part – Newton’s Law

5. 6.4 The Method of Sections
Members assumed to be in tension (BC
and GC) are subjected to a pull whereas
the member in compression (GF) is
subjected to a push
Apply equations of equilibrium

6. 6.4 The Method of Sections
Determining the Correct Sense of the
Unknown Member
Always assume the unknown member
forces in the cut section are in tension
- The numerical solution of the
equilibrium will yield positive scalars for
members in tension and negative
scalars for members in compression

7. 6.4 The Method of Sections
Determining the Correct Sense of the Unknown
Member
The correct sense of a direction of an unknown
force can be determined by inspection
- In more complicated problems, the sense of
the member can be assumed
- A positive answer indicates that the assumed
sense is correct and a negative answer indicates
that the assumed sense must be reversed

8. 6.4 The Method of Sections
Procedure for Analysis
Draw the FBD of a joint having at least one
known force and at most two unknown forces
If this joint is at one of the supports, it ma be
necessary to know the external reactions at the
truss support
Use one of two methods for determining the
correct sense of the member
Orient the x and y axes so that the forces on the
FBD can be easily resolved into x and y
components

9. 6.4 The Method of Sections
Procedure for Analysis
Free-Body Diagram
Decide how to cut or session the truss
through the members where forces are to
be determined
Before isolating the appropriate section,
determine the truss’s external reactions
Use the equilibrium equations to solve for
member forces at the cut session

10. 6.4 The Method of Sections
Procedure for Analysis
Free-Body Diagram
Draw the FBD of that part of the sectioned
truss which has the least number of forces
acting on it
Use one of the two methods for
establishing the sense of an unknown
member force

11. 6.4 The Method of Sections
Procedure for Analysis
Equations of Equilibrium
Moments are summed about a point that lies at
the intersection of lines of action of the two
unknown forces
The third unknown force is determined directly
from moment equation
If two of the unknown forces are parallel, forces
may be summed perpendicular to the direction of
these unknowns to determine the third unknown
force

12. 6.4 The Method of Sections
Example 6.5
Determine the force in members GE, GC,
and BC of the truss. Indicate whether the
members are in tension or compression.

13. 6.4 The Method of Sections
Solution
Choose section aa since it cuts through the
three members
FBD of the entire truss

14. 6.2 The Method of Joints
Solution
+ → ∑ Fx = 0;400 N − Ax = 0
Ax = 400 N
∑ M A = 0;
− 1200 N (8m) − 400 N (3m) + D y (12m) = 0
D y = 900 N
+ ↑ ∑ Fy = 0;
Ay − 1200 N + 900 N = 0
Ay = 300 N

15. 6.4 The Method of Sections
Solution
FBD of the sectioned truss

16. 6.2 The Method of Joints
Solution
∑ M G = 0;
− 300 N (4m) − 400 N (3m) + FBC (3m) = 0
FBC = 800 N (T )
∑ M C = 0;
− 300 N (8m) + FGE (3m) = 0
FGE = 800 N (C )
+ ↑ ∑ Fy = 0;
3
300 N − FGC = 0
5
FGC = 500 N (T )

17. 6.4 The Method of Sections
Example 6.6
Determine the force in member CF of the bridge
truss. Indicate whether the member are in tension
or compression. Assume each member is pin
connected.

18. 6.4 The Method of Sections
Solution
FBD of the entire truss

19. 6.4 The Method of Sections
Solution
FBD of the sectioned truss
Three unknown FFG, FCF, FCD

20. 6.4 The Method of Sections
Solution
Equations of Equilibrium
For location of O measured from E
4 / (4 + x) = 6 / (8 + x)
x = 4m
Principle of Transmissibility
∑ M O = 0;
− FCF sin 45o (12m) + (3kN )(8m) − (4.75kN )(4m) = 0
FCF = 0.589kN (C )

21. 6.4 The Method of Sections
Example 6.7
Determine the force in member EB of the roof
truss. Indicate whether the member are in
tension or compression.

22. 6.4 The Method of Sections
Solution
FBD of the sectioned truss

23. 6.4 The Method of Sections
Solution
Force system is concurrent
Sectioned FBD is same as the FBD for
the pin at E (method of joints)

24. 6.4 The Method of Sections
Solution
∑ M B = 0;
1000 N (4m) − 3000 N (2m) − 4000 N ( 4m) + FED sin 30o (4m) = 0
FED = 3000 N (C )
+ → ∑ Fx = 0;
FEF coso − 3000 cos 30o N = 0
FEF = 3000 N (C )
+ ↑ ∑ Fy = 0;
2(3000 sin 30o N ) − 1000 N − FEB = 0
FEB = 2000 N (T )